Maximum XOR Queries - Google Top Interview Questions


Problem Statement :


You are given a list of non-negative integers nums and a two-dimensional list of integers queries where each element contains [x, limit].

Return a list such that for each query [x, limit], we find an e in nums such that e ≤ limit and XOR(e, x) is maximized. If there's no such e, use -1.

Constraints

n ≤ 100,000 where n is the length of nums

m ≤ 100,000 where m is the length of queries

Example 1

Input

nums = [2, 4, 8]

queries = [

    [3, 5],

    [2, 0]

]

Output

[4, -1]

Explanation

For the first query, we can use 2 or 4 in nums. 2 ^ 3 = 1 while 4 ^ 3 = 7 so we pick 4 which yields the 

bigger XOR. In the second query, there's no number that's less than or equal to 0, so we set it to -1.


Solution :



title-img



                        Solution in C++ :

struct TrieNode {
    int lo = INT_MAX;
    TrieNode* children[2]{};
};

class Solution {
    public:
    vector<int> solve(vector<int>& nums, vector<vector<int>>& queries) {
        TrieNode* root = new TrieNode();
        for (int num : nums) {
            TrieNode* p = root;
            for (int i = 30; i >= 0; --i) {
                int nxt = (num & (1 << i)) ? 1 : 0;
                if (!p->children[nxt]) p->children[nxt] = new TrieNode();
                p = p->children[nxt];
                p->lo = min(p->lo, num);
            }
        }
        vector<int> ret;
        for (auto q : queries) {
            int x = q[0], limit = q[1];
            int ans = 0;
            TrieNode* p = root;
            for (int i = 30; i >= 0; --i) {
                if (x & (1 << i)) {
                    if (p->children[0]) {
                        p = p->children[0];
                    } else if (!p->children[1] || (p->children[1]->lo > limit)) {
                        ret.emplace_back(-1);
                        break;
                    } else {
                        p = p->children[1];
                        ans ^= (1 << i);
                    }
                } else {
                    if (p->children[1] && (p->children[1]->lo <= limit)) {
                        p = p->children[1];
                        ans ^= (1 << i);
                    } else if (!p->children[0]) {
                        ret.emplace_back(-1);
                        break;
                    } else {
                        p = p->children[0];
                    }
                }
                if (i == 0) ret.emplace_back(ans);
            }
        }
        return ret;
    }
};

vector<int> solve(vector<int>& nums, vector<vector<int>>& queries) {
    return (new Solution())->solve(nums, queries);
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(int[] nums, int[][] qs) {
        int n = nums.length;
        List<Map<Integer, Integer>> ls = new ArrayList<>();
        for (int i = 0; i < 32; i++) {
            Map<Integer, Integer> mp = new HashMap<>();
            for (int x : nums) {
                // set.add(x >> i);
                mp.put(x >> i, Math.min(x, mp.getOrDefault(x >> i, Integer.MAX_VALUE)));
            }
            ls.add(mp);
        }
        Arrays.sort(nums);

        int[] res = new int[qs.length];
        Arrays.fill(res, -1);
        if (n == 0)
            return res;
        int cnt = 0;
        for (int[] q : qs) {
            int v = q[0];
            int limit = q[1];
            int ans = 0;
            boolean find = false;
            for (int i = 31; i >= 0; i--) {
                Set<Integer> ts = ls.get(i).keySet();
                Map<Integer, Integer> mp = ls.get(i);
                // if(i == 0)System.out.println(ts);
                if ((v | (1 << i)) == v) { // this is one
                    int tk = ans >> i;
                    if (!ts.contains(tk)) {
                        ans |= 1 << i;
                    }

                    // System.out.println(tk+" "+i+" "+ts+" "+ans);
                } else { // we want to get index to be 1
                    int now = ans | (1 << i);
                    int tk = now >> i;
                    /// if we need this index but no that index
                    if (ts.contains(tk) && mp.get(tk) <= limit) {
                        ans = now;
                    }

                    // System.out.println(tk+" "+i+" "+ts+" "+ans);
                }
            }
            if (ans <= limit && ans >= nums[0]) {
                res[cnt] = ans;
            }
            cnt++;
        }
        return res;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, queries):
        nums.sort()

        def query(x, limit):
            start = 0
            stop = bisect_right(nums, limit)
            num = 0
            for i in range(31, -1, -1):
                i_th_bit = 1 << i
                plus = num + i_th_bit
                if nums[start] >= plus:
                    num = plus
                elif nums[stop - 1] >= plus:
                    cut = bisect_left(nums, plus, start, stop)
                    if x & i_th_bit:
                        stop = cut
                    else:
                        start = cut
                        num = plus

            return num

        return [query(x, limit) if nums and nums[0] <= limit else -1 for x, limit in queries]
                    

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